Best examples of Ksp examples: ion concentration in saturated solutions

Let’s begin the way you actually learn this topic: by looking at examples of Ksp examples: ion concentration in saturated solutions and building patterns from there.

When a sparingly soluble ionic solid sits in water, a tiny fraction dissolves until the solution is saturated. At that point, the ion product equals the solubility product constant, Ksp. From that single number and the dissolution equation, you can back out equilibrium ion concentrations.

Below, each real example of a Ksp problem follows the same playbook:

• Write the dissolution equation.
• Express ion concentrations in terms of the molar solubility, s.
• Plug into the Ksp expression.
• Solve for s, then get each ion concentration.

Once you’ve seen a few examples of Ksp examples: ion concentration in saturated solutions, you’ll start predicting the algebra before you even touch your calculator.

Example of a 1:1 salt: AgCl in pure water

Dissolution:

\[ \text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) \]

At 25 °C, a commonly cited value is:

\[ K_{sp}(\text{AgCl}) \approx 1.8 \times 10^{-10} \]

Let the molar solubility of AgCl be s mol/L. In a saturated solution:

• [Ag⁺] = s
• [Cl⁻] = s

So the Ksp expression becomes:

\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s \cdot s = s^2 \]

Solve for s:

\[ s = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \; \text{M} \]

So in this saturated solution:

• [Ag⁺] ≈ 1.3 × 10⁻⁵ M
• [Cl⁻] ≈ 1.3 × 10⁻⁵ M

This is one of the simplest examples of Ksp examples: ion concentration in saturated solutions: a 1:1 solid gives a quadratic in s that’s easy to solve by taking a square root.

If you want Ksp values straight from reference data, the NIST Chemistry WebBook is a solid starting point: https://webbook.nist.gov/chemistry/

1:2 salt example: CaF₂ and how stoichiometry amplifies anions

Now let’s look at calcium fluoride, where each formula unit produces one cation and two anions.

Dissolution:

\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\,\text{F}^-(aq) \]

Typical Ksp at 25 °C:

\[ K_{sp}(\text{CaF}_2) \approx 3.9 \times 10^{-11} \]

Let the molar solubility be s mol/L. In a saturated solution:

• [Ca²⁺] = s
• [F⁻] = 2s

Ksp expression:

\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s \cdot (2s)^2 = 4s^3 \]

Solve for s:

\[ s^3 = \frac{3.9 \times 10^{-11}}{4} = 9.75 \times 10^{-12} \]
\[ s = (9.75 \times 10^{-12})^{1/3} \approx 2.1 \times 10^{-4} \; \text{M} \]

Ion concentrations in the saturated solution:

• [Ca²⁺] ≈ 2.1 × 10⁻⁴ M
• [F⁻] ≈ 4.2 × 10⁻⁴ M

This example of a 1:2 salt highlights why you can’t just take the square root of Ksp every time. The stoichiometric coefficients become exponents in the Ksp expression, and they change the math from a square root to a cube root.

1:3 salt example: Fe(OH)₃ and very low solubility

Transition metal hydroxides are classic examples of Ksp examples: ion concentration in saturated solutions because they’re so sparingly soluble.

Dissolution:

\[ \text{Fe(OH)}_3(s) \rightleftharpoons \text{Fe}^{3+}(aq) + 3\,\text{OH}^-(aq) \]

A typical Ksp value at 25 °C is on the order of:

\[ K_{sp}(\text{Fe(OH)}_3) \approx 2.8 \times 10^{-39} \]

Let the molar solubility be s mol/L. Then:

• [Fe³⁺] = s
• [OH⁻] = 3s

Ksp expression:

\[ K_{sp} = [\text{Fe}^{3+}][\text{OH}^-]^3 = s \cdot (3s)^3 = 27s^4 \]

Solve for s:

\[ s^4 = \frac{2.8 \times 10^{-39}}{27} \approx 1.04 \times 10^{-40} \]
\[ s = (1.04 \times 10^{-40})^{1/4} \approx 1.0 \times 10^{-10} \; \text{M} \]

Ion concentrations:

• [Fe³⁺] ≈ 1.0 × 10⁻¹⁰ M
• [OH⁻] ≈ 3.0 × 10⁻¹⁰ M

This is a good reminder that some solids barely dissolve at all. But the same algebraic strategy still works: relate every ion concentration to s, plug into Ksp, solve.

For more background on solubility and precipitation in water systems, the USGS has a readable overview: https://www.usgs.gov/special-topics/water-science-school

Common-ion effect example: AgCl in 0.10 M NaCl

Now we’ll tweak the first example to show how a common ion slashes solubility.

Dissolution again:

\[ \text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) \]

Suppose the solution already contains 0.10 M NaCl. That means:

• Initial [Cl⁻] ≈ 0.10 M from NaCl

Let the additional solubility of AgCl be s mol/L. At equilibrium in this saturated solution:

• [Ag⁺] = s
• [Cl⁻] ≈ 0.10 + s ≈ 0.10 M (because s will be tiny compared to 0.10)

Ksp expression:

\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \approx s \cdot 0.10 \]

Solve for s:

\[ s = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9} \; \text{M} \]

Compare that to the earlier pure-water example of AgCl, where s was 1.3 × 10⁻⁵ M. The presence of a common ion (Cl⁻) cuts the solubility by about four orders of magnitude.

This is one of the best examples to remember: Ksp itself does not change with added NaCl, but the equilibrium concentrations do.

Environmental example: Pb(OH)₂ and drinking water pH

Ksp isn’t just a classroom exercise; it directly informs water treatment and public health.

Consider lead(II) hydroxide in water:

\[ \text{Pb(OH)}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{OH}^-(aq) \]

Typical Ksp:

\[ K_{sp}(\text{Pb(OH)}_2) \approx 1.2 \times 10^{-15} \]

Let’s say the water is adjusted to pH 11. That means:

• pH = 11 → pOH = 3 → [OH⁻] = 10⁻³ M

Now the Ksp expression is:

\[ K_{sp} = [\text{Pb}^{2+}][\text{OH}^-]^2 \]

Solve for [Pb²⁺]:

\[ [\text{Pb}^{2+}] = \frac{K_{sp}}{[\text{OH}^-]^2} = \frac{1.2 \times 10^{-15}}{(10^{-3})^2} = \frac{1.2 \times 10^{-15}}{10^{-6}} = 1.2 \times 10^{-9} \; \text{M} \]

So in a saturated solution at pH 11:

• [Pb²⁺] ≈ 1.2 × 10⁻⁹ M

This example of Ksp shows how raising pH (increasing [OH⁻]) forces dissolved lead to precipitate as Pb(OH)₂, lowering the free Pb²⁺ concentration. That logic underlies corrosion control and precipitation strategies used in drinking water systems.

For context on lead in drinking water and health limits, see the EPA overview: https://www.epa.gov/ground-water-and-drinking-water/basic-information-about-lead-drinking-water

Pharmaceutical-style example: Ca₃(PO₄)₂ and bioavailability

Calcium phosphate is a common ingredient in supplements and antacids, and it’s another good candidate for examples of Ksp examples: ion concentration in saturated solutions.

Dissolution:

\[ \text{Ca}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\,\text{Ca}^{2+}(aq) + 2\,\text{PO}_4^{3-}(aq) \]

Typical Ksp at 25 °C:

\[ K_{sp}(\text{Ca}_3(\text{PO}_4)_2) \approx 1.0 \times 10^{-26} \]

Let molar solubility be s mol/L. Then:

• [Ca²⁺] = 3s
• [PO₄³⁻] = 2s

Ksp expression:

\[ K_{sp} = [\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2 = (3s)^3(2s)^2 = 27s^3 \cdot 4s^2 = 108s^5 \]

Solve for s:

\[ s^5 = \frac{1.0 \times 10^{-26}}{108} \approx 9.3 \times 10^{-29} \]
\[ s = (9.3 \times 10^{-29})^{1/5} \approx 2.0 \times 10^{-6} \; \text{M} \]

Ion concentrations in this saturated solution:

• [Ca²⁺] ≈ 6.0 × 10⁻⁶ M
• [PO₄³⁻] ≈ 4.0 × 10⁻⁶ M

Real systems (like the human gut) are more complicated because phosphate can protonate (HPO₄²⁻, H₂PO₄⁻), but this idealized example of a Ksp calculation gives a feel for how little actually dissolves.

For a medically oriented look at calcium and phosphate balance in the body, NIH’s MedlinePlus is a good reference: https://medlineplus.gov/calcium.html

Mixed-solubility example: Predicting precipitation with Ksp

So far, our examples of Ksp examples: ion concentration in saturated solutions started with a solid and asked for ion concentrations. Now flip the direction: start from ion concentrations and ask whether a solid will precipitate.

Consider mixing equal volumes of:

• 0.010 M Pb(NO₃)₂ (source of Pb²⁺)
• 0.010 M NaCl (source of Cl⁻)

Right after mixing, both solutions are diluted by half, so initial concentrations:

• [Pb²⁺] = 0.0050 M
• [Cl⁻] = 0.0050 M

Possible precipitate: PbCl₂(s), with dissolution:

\[ \text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\,\text{Cl}^-(aq) \]

Typical Ksp:

\[ K_{sp}(\text{PbCl}_2) \approx 1.6 \times 10^{-5} \]

Compute the ion product, Q:

\[ Q = [\text{Pb}^{2+}][\text{Cl}^-]^2 = (0.0050)(0.0050)^2 = 0.0050 \times 2.5 \times 10^{-5} = 1.25 \times 10^{-7} \]

Compare Q to Ksp:

• Q = 1.25 × 10⁻⁷
• Ksp = 1.6 × 10⁻⁵

Since Q < Ksp, the solution is unsaturated with respect to PbCl₂. No precipitate forms; more PbCl₂ could dissolve if it were present. This is a slightly different but related example of using Ksp to reason about ion concentrations in a mixture.

2024–2025 context: Why Ksp still matters

If you’re wondering whether these examples of Ksp examples: ion concentration in saturated solutions are just old textbook artifacts, take a look at what’s happening in 2024–2025:

• Battery and materials research: Precipitation and solubility equilibria guide how solid electrolytes form and dissolve in next-gen batteries and fuel cells.
• Water treatment and PFAS removal: Engineers still rely on solubility and Ksp concepts to design precipitation steps that remove metals and other ions before advanced treatments.
• Biomineralization and medical imaging: Understanding how calcium phosphate, calcium carbonate, and contrast agents dissolve or precipitate in the body ties back to the same Ksp logic used in these classroom examples.

You won’t see Ksp splashed in headlines, but behind the scenes these real examples of solubility product calculations are part of how chemists make sure materials, medicines, and water systems behave the way we expect.

FAQ: Common questions about Ksp and ion concentrations

How do you recognize when to use Ksp in a problem?

You use Ksp whenever you’re dealing with a sparingly soluble ionic solid in equilibrium with its ions. If the problem mentions a “saturated solution,” “precipitation,” or asks whether a solid will form when solutions are mixed, it’s a strong signal you’ll be doing some version of these examples of Ksp examples: ion concentration in saturated solutions.

Can you give another quick example of a Ksp calculation?

Take BaSO₄, with dissolution:

\[ \text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]

If Ksp ≈ 1.1 × 10⁻¹⁰, set [Ba²⁺] = [SO₄²⁻] = s in a saturated solution. Then:

\[ K_{sp} = s^2 \Rightarrow s = \sqrt{1.1 \times 10^{-10}} \approx 1.0 \times 10^{-5} \; \text{M} \]

That’s another straightforward example of using Ksp to find ion concentrations.

Do temperature changes affect these Ksp examples?

Yes. Ksp values are temperature-dependent, and most tables list them at 25 °C. If the temperature changes, the Ksp changes, which shifts the equilibrium ion concentrations. The calculation steps stay the same, but you must use the Ksp value appropriate for that temperature.

How accurate are these examples compared with real lab data?

Real solutions have activity effects, side reactions (like complex ion formation), and pH-dependent speciation. The examples of Ksp examples: ion concentration in saturated solutions you see here assume ideal behavior and no side reactions. That’s usually good enough for classroom work and quick estimates. In research or industry, chemists use activity coefficients and more detailed models, but the core Ksp idea is identical.

If you can follow the logic in these examples of Ksp examples: ion concentration in saturated solutions, you already have the toolkit needed for most solubility and precipitation questions you’ll see in high school chemistry, general chemistry, or intro environmental science.